Pe Civil Eng Problems

44
1 Useful information, Formulas and Conversion Factors 8.34 pounds = 1 gallon of water 8.34 pounds/Million gallons = 1 mg/L 7.48 gallons = 1 cubic foot of water 325,851 gallons = 1 acre-foot of water 1440 minutes = one day 17.12 mg/liter = 1 grain per gallon 7000 grains = 1 pound 142.8 pounds/MG = 1 grain per gallon πD 2 = Area 4 πD = circumference of circle MW of carbon dioxide = 44 MW of calcium oxide = 56 MW of magnesium = 24.1 MW of calcium carbonate = 100 MW of CaCO3 = 100 MW of Ca = 40 MW of Mg = 24.1 MW of Na = 23 MW of S = 32 MW of O = 16 MW of F = 19 MW of Si = 28 Sodium Fluoride, NaF Sodium silicofluoride, Na2SiF6 Fluosilicic acid, H2SiF6 BV, cu. ft = Q/Q B , where design flow rate = Q; Q B = Breakthrough volume from an experimental pilot column M, lb = (BV)(ρ s ) where BV, cu. ft and ρ s = mass of carbon in lb/cu. ft. Mt = Q/Vb,, Mt = pounds exhausted per hour; Q = flow rate; Vb = gallons per pound treated BV, = area X length X (cu ft/ 1728 cu.in.)(7.48 gals/cu.ft)(3.785liters/gal) = liters Filter loading rate = gpm/sq.ft filter area Filter backwash rate = gpm/sq.ft.area BV = Q/Qb V = area X length Qb = (gal/hr)(BV/V)(cu.ft/7.48)=BV/hr. I=FQNE r /nE c where I= current in amperes, F=Faraday’s constant, E c = current efficiency as a fraction, Q = solution flow rate, N = normality of the solution, n = number of cells Power = RI 2 where R = resistance, I = current in amperes Faraday Constant = 96,500 ampere-seconds per gram equivalent

Transcript of Pe Civil Eng Problems

Page 1: Pe Civil Eng Problems

1

Useful information, Formulas and Conversion Factors

8.34 pounds = 1 gallon of water

8.34 pounds/Million gallons = 1 mg/L

7.48 gallons = 1 cubic foot of water

325,851 gallons = 1 acre-foot of water

1440 minutes = one day

17.12 mg/liter = 1 grain per gallon

7000 grains = 1 pound

142.8 pounds/MG = 1 grain per gallon

πD2 = Area

4

πD = circumference of circle

MW of carbon dioxide = 44

MW of calcium oxide = 56

MW of magnesium = 24.1

MW of calcium carbonate = 100

MW of CaCO3 = 100

MW of Ca = 40

MW of Mg = 24.1

MW of Na = 23

MW of S = 32

MW of O = 16

MW of F = 19

MW of Si = 28

Sodium Fluoride, NaF

Sodium silicofluoride, Na2SiF6

Fluosilicic acid, H2SiF6

BV, cu. ft = Q/QB , where design flow rate = Q; QB = Breakthrough volume from an

experimental pilot column

M, lb = (BV)(ρs) where BV, cu. ft and ρs = mass of carbon in lb/cu. ft.

Mt = Q/Vb,, Mt = pounds exhausted per hour; Q = flow rate; Vb = gallons per pound

treated

BV, = area X length X (cu ft/ 1728 cu.in.)(7.48 gals/cu.ft)(3.785liters/gal) = liters

Filter loading rate = gpm/sq.ft filter area

Filter backwash rate = gpm/sq.ft.area

BV = Q/Qb

V = area X length

Qb = (gal/hr)(BV/V)(cu.ft/7.48)=BV/hr.

I=FQNEr/nEc where I= current in amperes, F=Faraday’s constant, Ec = current efficiency

as a fraction, Q = solution flow rate, N = normality of the solution, n = number of cells

Power = RI2 where R = resistance, I = current in amperes

Faraday Constant = 96,500 ampere-seconds per gram equivalent

Page 2: Pe Civil Eng Problems

2

Probability equation of True recurrence interval: Z = 1 – (1-1/TR)N , where Z is the

probability of event will be equal or exceeded during a specified design period; TR is the

recurrence interval of the event; N is the design period in years

Kf = 528Qlog(r2/r1) / m(h2-h1), for artesian well where Kf is coefficient of permeability in

gpd/sq. ft.; Q = gallons per minute steady-state pumping rate; r2/r1 is drawdown in feet; m =

thickness of aquifer; h2 – h1 = 10 ft – 1 ft. = 9 feet

Kf = 1055Q log(r2/r1) / h22 – h1

2 , for unconfined aquifer

TR = V/Q where TR is the theoretical residence time of a process reator, V = volume; Q = flow,

Surface Overflow rate = gpd / surface area of tank, where gpd is gallons per day and

surface area is feet or gpd/ sq.ft.

Detention time = Volume of tank (in gallons) / flow rate (gallons/ time)

Weir loading, gpd/ft = flow, gpd / linear weir length, feet

T10 = C(tracer, mg/L) / Co(initial dose, mg/L) = 0.10

G = (P/µV)1/2

where G = velocity gradient, fps/ft; P = power input, ft-lb/s; µ = absolute

viscosity, lb . s/ft

2; V = volume, ft

3

Calculation for Volume of Grit Chamber or other reaction vessel: Volume, cu. ft. =

discharge rate, cfs X detention time, seconds

Calculation of length of Grit chamber or other reaction vessel: Length, feet = Volume, cu.

ft. / depth, feet

Calculation of cross sectional area of grit chamber or other reaction vessel:

Cross sectional area = cfs/ fps, where cfs is cubic feet per second and fps is feet per second

Filtration rate, gpm/sq. foot = gpm/area, sq. feet

Kozeny Equation for rate of head loss thru clean granular medium:

h/l = Jv(1-ε)2VS

2 / gε

3d

2 where h/l = head loss per unit depth of filter bed, gt H2O/ft

J = constant, approximately 6 for filtration in the laminar flow region, dimensionless

v= kinematic viscosity (µ/p), ft2/s, (m

2/s)

g = acceleration of gravity, ft/s2

ε = porosity of the stationary filter bed, dimensionless

V = superficial (approach) velocity of water above the bed, ft/s

S = shape factor ranging from 6.0 for spherical grains to 7.5 for angular grains, dimensionless

d = mean grain diameter, ft.

Calculate Chemical Dosage: Million gallons X 8.34lbs/million gallons X ppm = pounds.

Calculate the meq/l of water anions and cations of a water analysis:

Meq/L = (Valence / MW) X mg/L or Meq/L = (1 / Eq.Wt.) X mg/L

T. Hardness, as CaCO3 = (Ca as Ca, mg/L X 2.5) + (Mg as Mg, mg/L X 4.12)

The calcium alkalinity = Calcium hardness or total alkalinity whichever is smaller

Magnesium alkalinity = magnesium hardness if total alkalinity ≥ than total hardness

Magnesium Alkalinity = Total Alkalinity – calcium hardness if total alkalinity is > than

calcium hardness but less than total hardness.

Sodium alkalinity = total alkalinity – total hardness

NCH = Total Hardness – Total Alkalinity ( If Mg Alkalinity present then no Ca NCH)

Page 3: Pe Civil Eng Problems

3

Water

1. Groundwater is used for a water supply. It is taken from the ground at 25 degrees

C. The initial properties of the water are as follows.

CO2 = 60 mg/L as CaCO3

Alkalinity = 200 mg/L as CaCO3

pH = 7.1

The water is treated for CO2 removal by spraying into the atmosphere through a

nozzle. The final CO2 concentration is 5.6 mg/L as CaCO3 at 25 degrees C. The

first ionization constant of carbonic acid is 4.45 X 10-7

. What is the final pH of the

water after spraying and recovery, assuming the alkalinity is unchanged in the

process? Since pH < 8.3, all alkalinity is in the bicarbonate (HCO3

-) form. Therefore, the equilibrium

expression for the first ionization of carbonic acid may be used.

CO2 + H2O → H2CO3 ↔ H+ + HCO3

-

K1 = [H+][HCO3

-] = 4.45 X 10

-7

[H2CO3]

The coefficients of CO2 and H2CO3 are 1. The number of moles of each compound is the same.

One mole of CO2 produces one mole of H2CO3.

[H2CO3- = [CO2] = 5.6 mg/L / (100 g/mol)(1000 mg/g) = 5.6 X 10

-5

[HCO3-] = 200 mg/L / 100 g/mol (1000 mg/g) = 1.246 X 10

-8

Solve for the hydrogen ion concentration.

[H+] = K1(H2CO3] = K1[H2CO3] = (4.45 X 10

-7)(5.6 X 10

-8) = 1.246 X 10

-8)

[HCO3-] [CO2]

pH = -log[H+] = - log(1.246 X 10

-8) = 7.9

2. The solids concentration of a stream water sample is to be determined. The total

solids concentration is determined by placing a portion of the sample into a porcelain

evaporating dish, drying the sample at 105 degrees C, and igniting the residue by placing

the dried sample in a muffle furnace at 550 degrees C. The following masses are recorded.

Mass of empty dish: 50.326 g

Mass of dish and sample: 118.400 g

Mass of dish and dry solids: 50.437 g

Mass of dish and ignited solids: 50.383 g

a). The total solids concentration is

The density of water is 1 g/mL. The volume of the tested sample is

118.4 g – 50.326 g / 1 g/mL = 68.1 mL

TS = (50.437 g – 50.326 g) (1000 mg/L) (1000 mg/L) = 1630 mg/L

68.1 ml

b). The total volatile solids concentration is:

TVS = (50.437 g – 50.383 g) X (1000 mg/g (1000 ml/L) = 793 mg/L

68.1 ml

c). The total fixed solids concentration is

TFS = 1630 mg/L – 793 mg/L = 837 mg/L

Page 4: Pe Civil Eng Problems

4

3. The dissolved concentration is determined by filtering a portion of the sample

through a glass-fiber disk into a porcelain evaporating dish, drying the sample at 105

degrees C, and igniting the residue by placing the dried sample in a muffle furnace at 550

degrees C. The following masses are recorded.

The following masses are recorded.

Volume of sample = 25 mL.

Mass of empty dish: 51.494 g

Mass of dish and dry solids = 51.524 g

Mass of dish and ignited solids = 51.506 g

a. The total dissolved solids concentration is:

TDS = (51.524 g) – (51.494 g) X (1000 mg/g)(1000mg/L) = 1200 mg/L

25 ml

b. VDS = (51.524 g – 51.506 g) (1000 mg/L)(1000 mL/L = 720 mg/L

25 ml

c. FDS = 1200 mg/L – 720 mg/L = 480 mg/L

Clarifiers

4. What is the theoretical mean residence time of a process reactor if the volume of the

reactor is 100,000 gallons and the rate of flow through the reactor is 1000 gpm.

Solution:

TR = V/Q where V = volume; Q = flow

TR = 100000 gallons / 1000 gpm = 100 minutes or 1.67 hours

5. Calculate the overflow rate of a rectangular clarifier that is 90 feet long, 16 feet wide,

and 12 feet deep with a flow rate of 1.5 mgd.

Solution: Overflow rate = gpd / surface area of tank = 1,500,000 gallons / 90 feet long X 16

feet wide = 1041 gpd/sq.ft.

Page 5: Pe Civil Eng Problems

5

6. Calculate the detention time in hours of a rectangular clarifier that is 90 feet long, 16

feet wide, and 12 feet deep with a flow rate of 1.5 mgd.

Detention time = Volume of tank (in gallons) / flow rate (gallons/ time)

Detention time, hours = 90 feet X 16 feet X 12 feet X 7.48gals/cu. Ft / 1500000 gpd / 24 hours or

62500 gallons per hour = 2.1 hours.

7. Calculate the weir loading for a rectangular clarifier that is 90 feet long 16 feet wide,

and 12 feet deep that treats 500, 000 gallons per day with a weir length equal to three tank

widths.

Solution:

Weir loading, gpd/ft = flow, gpd / linear weir length, feet

Weir loading, gpd/ft = 500000 gpd / 3 X 16 ft = 10416 gpd/foot

8. A completely mixed reactor showed the following results of a step-dose input tracer test

using fluoride chemical. Calculate the T10 in minutes. The T10 time in minutes is when

10% of the tracer chemical passes through the effluent of the mixed reactor.

Solution: T10 = C(tracer, mg/L) / Co(initial dose, mg/L) = 0.10. The T10 will fall between 12

and 15 minutes. By interpolation, T10 = 0.045/0.1 X 3 minutes difference + 12 minutes = 13.35

minutes

9. A water plant has a flocculation tank 64 feet long and 100 feet wide with a water depth

of 16 feet and a power input of 3960 foot lbs/ s or 5.36 horsepower. Calculate the velocity

gradient if µ = 2.735 X 10-5

lb-s/ft2.

Solution: G = (P/µV)1/2

where G = velocity gradient(power to the water), fps/ft; P = power

input, ft-lb/s; µ = absolute viscosity, lb . s/ft

2; V = volume, ft

3

Page 6: Pe Civil Eng Problems

6

G = (3960 / 2.735 X 10-5

X 16 X 64 X 100) ½

= 38 fps/ft

10. Determine the volume of a hopper-bottom grit removal tank if the detention time is

1 minute at 1 cfs peak flow.

Solution: Volume, cu. ft. = discharge rate, cfs X detention time, seconds

Volume = 1 cfs X 60 seconds = 60 cubic feet.

11. Determine the length of a grit chamber if the detention time is 1 minute at peak flow

of 1 cfs with flow velocity of 1 fps and depth of chamber is 2 feet.

Solution:

Q, cfs X detention time, seconds = Volume, cubic feet

1 cfs X 60s = 60 cubic foot

Length, feet = Volume, cu. ft. / depth, feet = 60 cu.ft. / 2 foot depth = 30 feet long

12. Determine the cross-sectional area of a grit chamber if the detention time is 1

minute at peak flow of 1 cfs with flow velocity of 1 fps.

Solution: The cross sectional area is equal to the discharge rate divided by the velocity.

Cross sectional area = cfs/ fps = 1 cfs / 1 fps = 1 sq. foot

13. Determine the gpm/square foot filtration rate of a filter 26 feet by 26 feet that treats

4 mgd.

Solution:

Filtration rate, gpm/sq. foot = gpm/area, sq. feet = 4,000,000/1440 min/day / 26 X 26 = 4.1

gpm/sq.ft.

14. Calculate the initial head loss through a filter with 18 inches of uniform sand having

a porosity of 0.42 and a grain diameter of 1.6 X 10-3

ft. Assume spherical particles and a

water temperature of 50 degrees F. The filtration rate is 2.5 gpm/sq.ft. Use the Kozeny

equation.

h/l = Jv(1-ε)2VS

2 / gε

3d

2 where h/l = head loss per unit depth of filter bed, gt H2O/ft

J = constant, approximately 6 for filtration in the laminar flow region, dimensionless

v= kinematic viscosity (µ/p), ft2/s, (m

2/s)

g = acceleration of gravity, ft/s2

ε = porosity of the stationary filter bed, dimensionless

V = superficial (approach) velocity of water above the bed, ft/s

S = shape factor ranging from 6.0 for spherical grains to 7.5 for angular grains, dimensionless

d = mean grain diameter, ft.

V = 2.5 gpm/sq.ft / 7.48 gal/ft X 60 sec/min = 0.00557 ft/sec, 1 = 1.5 ft.

h/l = 6 X 1.410 X 10-5

(1 – 0.42)20.00557 X (6.0)

2 / 32.2 X (0.42)

3 X (1.6 X 10

-3)2 = 0.93 feet

h = 1.5 ft depth X 0.933 = 1.4 feet loss of head.

Based on a given chemical dosage in ppm, calculate the pounds required per million

gallons.

Solution:

Million gallons X 8.34lbs/million gallons X ppm = pounds.

Example: Calculate how many pounds per day of chlorine if the dosage is 3.5 ppm and the

amount of water to be treated is 10 million gallons per day.

Answer: 10 MGD X 8.34 X 3.5 ppm = 292 pounds per day.

Page 7: Pe Civil Eng Problems

7

15. Calculate the meq/l of water anions and cations of a water analysis.

Meq/L = (Valence / MW) X mg/L or Meq/L = (1 / Eq.Wt.) X mg/L

Table 11.3 - Example of Water Analysis Data

Ion Mg/L Equivalent weight Meq/l

Ca2+

40 20 2.00

Mg2+

10 12.2 0.82

Na+

11.7 23.0 0.51

K+

7.0 39.1 0.18

Total Cations = 3.51

HCO3-

110 61.0 1.80

SO42-

67.2 48.0 1.40

Cl-

11.0 35.5 0.31

Total Anions= 3.51

Name the three major classes of alkalinity

The three major classes of alkalinity are defined by their pH range.

a. Bicarbonates, HCO3- >4.5 to ≤ pH 8.3

b. Carbonates, CO3 - >8.3 < pH 10

c. Hydroxide, OH- >pH10

What is the titration endpoint for phenolthalein alkalinity and the total alkalinity

respectively.

P. Alkalinity = pH 8.3

T. Alkalinity = 4.5

16. Determine the alkalinity relationships from a water analysis giving the phenol

alkalinity and total alkalinity concentrations.

Titration Result OH-

Alkalinity as CaCO3

CO3-

Alkalinity as CaCO3

HCO3-

Alkalinity as

CaCO3

P = 0 0 0 T

P<1/2 T 0 2P T-2P

P=1/2 T 0 2P 0

P>1/2 T 2P – T 2(T- P) 0

P = T T 0 0

17. Fill in the correct amount in mg/L for each blank based on alkalinity results in the

first column.

Alkalinity Tests

Results

OH- Alkalinity as

CaCO3

CO3- Alkalinity as

CaCO3

HCO3- Alkalinity as

CaCO3

P.Alk = 0, T. Alk. =100 0 0 100

P Alk = 40, T.Alk.=100 0 80 20

P Alk = 50, T.Alk. =100 0 100 0

P Alk=60, T Alk.=100 20 80 0

P = 100, T Alk.=100 100 0 0

Page 8: Pe Civil Eng Problems

8

18. What are the atomic weight and the equivalent weight of Na+1

and Ca+2

if a water

sample has 102 mg/l of Na+1

and 68 mg/L of Ca+2

, how many milliequivalents of each are

present?

Na+

: AW = 23, Eq. Wt. = 23/1 = 23 grams, 23 mg = meq.

Meqs = (102 mg/L) / 23 mg/meq = 4.43 meqs/L

Ca+2

: AW = 40, Eq. Wt. = 40/2 = 20 grams, 20 mg = meq.

Meqs = (68 mg/L) / 20 mg/meq. = 3.4 meqs/L

19. What are the molecular weight and the equivalent weight of CaCl2 if a water sample

contains 168 mg/L of CaCl2, how many milliequivalents are present?

CaCl2 MW = 40 + 2(35.45) = 110.45

Meq. Wt.= 110/2 = 55 mg

Meqs = (168 mg/L) / 55 mg/meq = 3.05 meg/L

20. If a water sample contains 134 ppm of Na+1

, how many mg/L are present?

1 ppm = 1 mg/L if specific gravity = 1.00, therefore 134 ppm = 134 mg/L.

21. If a water sample has a pH of 7.6, how many moles per liter of hydrogen ion, H+,

and hydroxyl ion, OH-, are present?

pH = 7.6

pH = Log 1/[H+]

[H+] = 10

-7.6 moles /L

[H+][OH

-] = 10

-14

[10-7.6

][OH-] = 10

-14

[OH-] = 10

-6.4 moles/L

In water coagulation with alum (aluminum sulfate) and bicarbonate alkalinity, the

simplified reaction is: Al2(SO4)3 . 14 H2O + 6 HCO3 → 2Al(OH)3 ↓ 3SO4

-2 + 14 H2O +

6CO2, why does this reaction go to completion?

It goes to completion because an insoluble product is formed.

22. Determine the total hardness as calcium carbonate if given the calcium and

magnesium concentrations.

Solution: T. Hardness, as CaCO3 = (Ca as Ca, mg/L X 2.5) + (Mg as Mg, mg/L X 4.12)

Example: Ca = 30 mg/L as Ca X 2.5 = 75 mg/L as CaCO3

Mg = 10 mg/L as Mg X 4.12 = 41.2 mg/L as CaCO3

Total Hardness as CaCO3 = 116.2 mg/L as CaCO3

Name three ways carbonate hardness can be reduced in water.

Boiling, lime, and ion exchange

Name two ways noncarbonated hardness can be reduced in water.

Soda ash and ion exchange.

Write the chemical reactions in lime treatment of carbon dioxide. CO2 + Ca(OH)2 CaCO3 + H2O

Write the chemical reactions in lime treatment of calcium and magnesium bicarbonate.

Ca + 2HCO3 + Ca(OH)2 2CaCO3 + 2H2O(pH 8.3-9.)

Mg + 2HCO3 + Ca(OH)2 CaCO3 + Mg + CO3 + 2H2O(pH >10.8)

Write the chemical reactions in lime treatment of magnesium carbonate and sulfate.

Mg + CO3 + Ca(OH)2 → CaCO3 + Mg(OH)2

Mg + SO4 + Ca(OH)2 Ca + SO4 + Mg(OH)2

Page 9: Pe Civil Eng Problems

9

Write the chemical reactions of the second state treatment with soda ash of non-carbonate

hardness.

Ca + SO4 + Na2CO3 Na2SO4 + CaCO3

Ca + Cl2 + Na2CO3 CaCO3 + 2NaCl

23. A water has an hardness of 185 m/L as CaCO3, what is the hardness expressed as

meq/L?

CaCO3 Meq wt = [40 +12+3(16)] / 2 = 50

Meqs/L = (185 mg/L) / 50 mg/meq. = 3.70 meq/L

25. A water has an alkalinity of 225 mg/L as CaCO3. What is the alkalinity expressed as

meq/L?

CaCO3 meq wt. = [40+12+3(26)] / 2 = 50

Meq/L = (225 mg/L) / 50 mg/meq = 4.5 meq.

How can the calcium alkalinity be determined?

The calcium alkalinity = Calcium hardness or total alkalinity whichever is smaller

Name two ways the magnesium alkalinity be determined?

Magnesium alkalinity = magnesium hardness if total alkalinity ≥ than total hardness

Magnesium Alkalinity = Total Alkalinity – calcium hardness if total alkalinity is > than

calcium hardness but less than total hardness.

How can the sodium alkalinity be determined?

Sodium alkalinity = total alkalinity – total hardness

How can the noncarbonated hardness be determined? NCH = Total Hardness – Total Alkalinity ( If Mg Alkalinity present then no Ca NCH)

26. Determine the calcium alkalinity, magnesium alkalinity, sodium alkalinity, calcium

non-carbonate hardness, and the magnesium non-carbonate hardness from the given water

analyses.

AAnnaallyysseess WWaatteerr ##11 WWaatteerr ##22 WWaatteerr ##33

TToottaall HHaarrddnneessss 330000 330000 330000

CCaallcciiuumm HHaarrddnneessss 220000 220000 220000

MMgg HHaarrddnneessss 110000 110000 110000

TToottaall AAllkkaalliinniittyy 115500 225500 335500

IInntteerrpprreettaattiioonnss WWaatteerr ##11 WWaatteerr ##22 WWaatteerr ##33

CCaallcciiuumm AAllkkaalliinniittyy 115500 220000 220000

MMgg.. AAllkkaalliinniittyy NNoonnee 5500 110000

SSooddiiuumm AAllkkaalliinniittyy NNoonnee NNoonnee 5500

CCaa NN..CC.. HHaarrddnneessss 5500 NNoonnee NNoonnee

MMgg.. NN..CC.. HHaarrddnneessss 110000 5500 nnoonnee

What is the practical minimum total hardness level of the lime-soda ash treatment method?

The practical minimum hardness level using the lime-soda ash treatment method is 50 mg/L.

Page 10: Pe Civil Eng Problems

10

27. Calculate the hydrated lime (100%), soda ash, and carbon dioxide requirements to

reduce the hardness of a water with the following analysis to about 50 to 80 mg/L by the

excess Lime-soda ash process based on the following water analyses.

Analyses: Total Hardness = 280 mg/L as CaCO3

Mg++ = 21 mg/L

Alkalinity = 170 mg/L as CaCO3

Carbon Dioxide = 6 mg/L

Lime Requirement: Carbon Dioxide = (6) (56) / (44) = 8

Alkalinity = (170) (56) / (100) = 95

Mg ++ = (21) (56) / (24.3) = 48

Excess Lime = = 35

Total CaO required = 186 mg/L

Soda Ash Requirement: NCH = 280 – 170 = 110 mg/L

Soda Ash (Na2CO3) = (110) (106) / (100) = 117 mg/L

28. A town’s water supply has the following ionic concentrations.

Al+++

= 0.5 mg/L

Ca++

= 80.2 mg/L

Cl- = 85.9 mg/L

CO2 = 19 mg/L

CO3-- = 0

Fe++

= 1.0 mg/L

Fl- = 0

HCO3- = 185 mg/L

Mg++

= 24.3 mg/L

Na+ = 46.0 mg/L

NO3- = 0

SO4-- = 125 mg/L

a). What is the total hardness?

Ca++

= 80.2 mg/L X 2.5 = 200.5 mg/L as CaCO3

Mg++

= 24.3 mg/L X 4.12 = 100 mg/L

Fe ++ = 1.0 mg/L X 1.79 = 1.79 mg/L

Al+++

= 0.5 mg/L X 5.56 = 2.78 mg/L

Total Hardness ------------ 305 mg/L

b). How much slaked lime is required to combine with the carbonate hardness?

To remove the carbonate hardness, CO2: 19 mg/L X 100/44 or 2.27 = 43.13 mg/L as CaCO3

HCO3- : 185 mg/L X 50mg/Las 1 meq/L/61mg/L as 1meq/L or 0.82 = 151.7 mg/L

Total equivalents to be neutralized are: 43.13 mg/L + 151.7 mg/L = 194.83 mg/L as Ca(OH)2

Convert Ca(OH)2 to CaCO3 = 194.83 /100/74 or 1.35 = 144.3 mg/L as CaCO3 as substance

c). How much soda ash is required to react with the non-carbonate hardness?

NONE, because soda ash only reacts with non-carbonate hardness and the problem is only

asking for the carbonate hardness removal which is only removed by lime. However, there is

non-carbonate hardness present in the water. It is calculated by the difference in the Total

hardness and the total alkalinity or 305 mg/l – 152 mg/L = 153 mg/L as CaCO3 of non-carbonate

hardness.

Page 11: Pe Civil Eng Problems

11

29. If the town’s water supply has the following ionic concentrations and it is to be

softened using a zeolite process with the following characteristics: exchange capacity,

10,000 grains/cu.ft., and salt requirement , 0.5 lbm per 1000 grains hardness removed.

How much salt is required to soften the water to 100 mg/L hardness?

Ca(HCO3)2 = 137 mg/L as CaCO3

MgSO4 = 72 mg/L as CaCO3

CO2 = 0

There are 7000 grains in a pound. The hardness removed is:

137 mg/L + 72 mg/L – 100 mg/L = 109 mg/L

(109 mg/L) (8.345 lbm-L/mg-MG) = 909.6 lbm hardness/MG

(0.5 lbm/1000 gr)(909.6 lbm/MG)(7000 gr/lbm) = 3.18 X 103 lbm/MG or 3200 lbm/MG

30. Determine the % hypochlorous acid (HOCl) available for disinfection from chlorine at

7.0, 8.0, and 9.0 pH values at 20 degrees C.

Solution: At pH 7.0 and 20 degrees C, the % hypochlorous acid is 80%. At pH 8.0 and 20

degrees C, the % hypochlorous acid is 30%. At pH 9.0 and 20 degrees C, the % hypochlorous

acid is 5%.

Page 12: Pe Civil Eng Problems

12

31. Determine the storage requirement and critical design period of a

reservoir based on the given mean inflow and draft flows for a year period. Col 1 Col 2 Col 3 Col 4 Col. 5 Col. 6

Month Inflow

MG

Draft

MG

Sum of

Inflows (MG)

Defieciency

MG

Cumulative

Deficiency(MG)

Jan 37.2 53.1 37.2 15.9 15.9

Feb 64.8 53.1 102 -11.7 0

Mar 108 53.1 210 -54.9 0

Apr 12 53.1 222 41.1 41.1

May 8.4 53.1 230.4 44.7 85.8

June 9.6 53.1 240 43.5 129.3

July 2.4 53.1 242.4 50.7 180

Aug 33.6 53.1 276 19.5 199.5

Sept 50.4 53.1 326.4 2.7 202.2

Oct 129.6 53.1 456 -76.5 0

Nov 117.6 53.1 573.6 -64.5 0

Dec. 26.4 53.1 600 26.7 26.7

Solution: Col 4 = Sum of Col 2; Col 5 = Col 3 – Col 1; Col 6 = Sum of Col 5 when > 0

Answer: Storage Requirement = 202.2 MG; Critical Design Period = 202.2/col 3 or 53.1=3.8

months.

32. Write the probability equation of a true recurrence interval and develop a

probability table based on the recurrence intervals of 1 year, 10 years, 50 years and 100

years for design periods of 1 year, 10 years 50 years and 100 years.

Answer: Z = 1 – (1-1/TR)N , where Z is the probability of event will be equal or exceeded

during a specified design period; TR is the recurrence interval of the event; N is the

design period in years

Example: Z = 1 – (1- 1/10)10

= 1 – (1-0.1)10

= 1 – (0.9)10

= 1 – (1 – 0.35) = 0.65

Table : Probability That an Event Having a Prescribed Interval Will Be Equaled

or Exceeded During a Specified Design Period

33. Design Period (Years)

TR (year) 1 10 50 100

1 1.0 1.0 1.0 1.0

10 0.1 0.65 0.995 ~1.0

50 0.02 0.18 0.64 0.87

100 0.01 0.10 0.40 0.63

Page 13: Pe Civil Eng Problems

13

33. A flow of 100 mgd is to be developed from a 190 sq. mile watershed. At the flow line

the area’s reservoir is estimated to cover 3900 acres. The annual rainfall is 40 inches, the

annual runoff is 14 inches, and the annual evaporation is 49 inches. Find the net gain or

loss in storage this represents. Calculate the volume of water evaporated in acre-ft.

Solution:

Reservoir area, square miles = 3900acres / 640 acres per square mile = 6.1 square miles

Annual runoff, ac-ft. = (14inches annual runoff / 12inches per foot)(190 sq. miles watershed –

6.1 sq. miles for reservoir area) (640 acres per sq. mile reservoir area) = 137,704 ac-ft.

Annual evaporation, ac-ft. = (49 inches per year/12 inches per foot)(3900 acres reservoir area)

= 15,925 ac-ft.

Draft, ac-ft. = (100 MGD X 365 days X 1000000 gallons per Million) / (7.48 gallons per cubic

foot X 43,560 square feet per acre) = 112,022 ac-ft.

Precipitation on Lake, ac-ft. = (40 inches per year / 12 inches per foot) (3900 acres in

Reservoir area) = 13,000 ac-ft.

Gain in Storage, ac-ft. = 137,704 ac-ft runoff + 13,000 ac-ft precipitation in lake = 150,704 ac-

ft.

Loss in Storage, ac-ft. = 112,022 ac-ft Draft + 15925 ac-ft annual evaporation = 127,947 ac-ft

loss in storage.

Net gain in Storage, ac.-ft. = 150,704 ac-ft – 127,947 ac-ft = 22,757 ac-ft. gain in storage

34. If a constant annual yield of 1500 gpm was required, what reservoir capacity would

be needed to sustain it? Fine the capacity in acre-ft per year.

Solution:

Constant annual yield, gpm = 1500 gpm

Time of operation without recharge = 1 year

Reservoir Capacity Required, ac.-ft. = 1500 gpm X365 days per year X24 hours per day X 60

minues per hour /325851gallons / ac.ft. = 788,000,000 gallons/325851 gallons per ac-ft = 2418

ac.ft

35. Determine the permeability of an artesian aquifer being pumped by a fully

penetrating well. The aquifer is composed of medium sand and is 90 feet thick. The

steady-state pumping rate is 850 gpm. The drawdown of an observation well 50 feet away

is 10 feet, and the drawdown in a second observation well 500 feet away is 1 foot.

Solution:

Use Coefficient of Permeability Equation below:

Page 14: Pe Civil Eng Problems

14

Kf = 528Qlog(r2/r1) / m(h2-h1) = 528 X 850 X log (10/1) / 90 X (10 – 1) = 554 gpd/sq. foot where Kf is coefficient of permeability in gpd/sq. ft.; Q = gallons per minute steady-state

pumping rate; r2/r1 is drawdown in feet = 500/50 = 10 feet; m = 90 feet thickness of aquifer; h2

– h1 = 10 ft – 1 ft. = 9 feet

36. An 18-inch well fully penetrates an unconfined aquifer of 100 feet depth. Two

observation wells located 100 and 235 feet from the pumped well have drawdowns of 22.2

and 21 feet, respectively. If the flow is steady and Kf = 1320 gpd per square feet, what

would be the discharge?

Equation: Kf = 1055Q log(r2/r1) / h22 – h1

2 , for unconfined aquifer

Solve for Q = Kf (h22 – h1

2) / 1055 log(r2/r1)

Log(r2/r1) = log (235/100) = 0.37107

h2 = 100 – 21 = 79 feet

h1 = 100 – 22,2 = 77.8 feet

Q = 1320 ( 792 – 77.8

2) / 1055 X 0.37107 = 634.44 gpm

37. Compare the annual water requirements of an 1500-acre irrigated farm and a city

of 130000 population. Assume an irrigation requirement of 3 acre-feet/acre/year and a per

capita water use rate of 180 gpcd.

City Use, gpd = 130000 population X 180 gpcd = 2,340,000 gallons per day

Irrigation use, gpd = 43560 sq. ft per acre X 7.48 gallons / cu.ft. X 3 acre-ft X 1500 acres / 365

days per year = 4,017,067 gpd

2,340,000 gpd / 4,017,067 gpd = to the Irrigation use of 1.72 times the city’s use.

38. Consider a 1000-acre residential area with a housing density of four dwellings per

acre. Estimate the peak hourly water use requirement.

Page 15: Pe Civil Eng Problems

15

Number of dwelling units = 1000 X 4 = 4000 units

Use Figure 4.4 and read up from y-axis = 4000 units

Find curve line for 4.0 dwelling units per acre that crosses the 4000 unit line at 5000 gpm as

peak hourly flow.

39. A community had a population of 200,000 in 2000 and it is expected that this will

increase to 260,000 by 2015. The water treatment capacity in 2000 was 43 mgd. A survey

showed that the average per capita water use rate was 180 gpcd. Estimate the community’s

water requirements in 2015 assuming (a) no change in use rate and (b) a reduced rate of

160 gpcd. Will expanded treatment facilities be needed by 2015 for condition (a)? For

condition (b)?

Solution:

In 1995 the population is 200,000 with water treatment capacity of 43 mgd with 180 gpcd water

use.

In 2010 the population will be 260,000.

(a) At no change in use rate of 180 gpcd, the water treatment capacity will be how much at

260000 population: 260000 X 180 = 46.8 mgd, expansion is required

(b) If the per capita use decreases to 160 gpcd by 2015, how much water treatment capacity will

be needed? 260000 X 160 = 41.6 mgd, less capacity required

Page 16: Pe Civil Eng Problems

16

Wastewater

Wastewater Terminolgy

Define the following:

Bacteria – are the simplest forms of microorganisms that use soluble food and are capable of

self-reproduction.

Heterotrophic bacteria – use organic compounds as an energy and carbon source for synthesis.

Saprophte – refers to an organism that lives on dead or decaying organic matter.

Aerobes – require free dissolved oxygen to live and multiply.

Anaerobes – oxidize organic matter in the complete absence of dissolved oxygen.

Facultative bacteria – are a class of bacteria that use free dissolved oxygen when available but

can also respire and multiply in its absence.

Escherichia coli – is a fecal coliform that is a facultative bacterium.

Autotrophic bacteria – use carbon dioxide as a carbon source and oxidize inorganic compounds

for energy.

Thiobacillus – are autotrophic sulfur bacteria that convert hydrogen sulfide to sulfuric acid.

Leptothrix and Crenothrix – are iron-accumulating bacteria and thrive in water pipes containing

dissolved iron and form yellow-or reddish-colored slimes.

Fungi – refer to microscopic non-photosynthetic plants, including yeasts and molds.

Algae – are autotrophic microscopic photosynthetic plants that uses carbon dioxide or

bicarbonates in solution as a carbon source with phosphorous and nitrogen necessary for growth.

Photosynthesis – is illustrated by the equation:

sunlight

CO2 + 2H2O new cell tissue + O2 +H2O)

Dark reaction

Protozoans – are single-celled aerobes that reproduce by binary fission and ingest bacteria and

algae.

Rotifers – are the simplest multicelled animals (aerobes) found in low pollutional streams and

lakes.

Daphnia and Cyclops - are crustaceans which are strict aerobes that ingest microscopic plants

and serve as food for fishes.

Metabolism – is the biochemical process performed by living organisms to yield energy for

synthesis, motility and respiration to remain viable.

Oxidation – is the addition of oxygen, removal of hydrogen, or removal of electrons.

Reduction – is the removal of oxygen, addition of hydrogen, or addition of electrons.

Exponential growth phase – is limited only by the microorganisms ability to process the

substrate, i.e., excess food and maximum rate of metabolism.

Declining growth phase – is caused by an increasing shortage of substrate.

Stationary Phase – The rate of reproduction of microorganisms equals the rate of death of the

microorganisms.

Endogenous Growth Phase – of microorganisms occurs when the rate of metabolism is

decreasing at an increasing rate, resulting in a rapid decrease in the number of viable cells.

Food to microorganism ratio (F/M) – is the balance between the food supply and mass of

microorganisms.

Page 17: Pe Civil Eng Problems

17

1. Compare the average daily sewage flow from an apartment building of 15 floors

having 10 apartments per floor using 200 gpd/unit with the sewage flow from a 10-acre

residential area having four houses per acre with an average of 4 persons per house using

90 gpcd.

For individual houses there will be:

10 acre residential area X 4 houses per acre = 40 houses per 10 acres

4 persons per house = 40 houses per 10 acres X 4 persons per house = 160 persons

160 persons X 90 gpcd = 14,400 gpd

For the apartment there will be:

10 apartments per floor X 15 floors = 150 units

150 units @ 200 gpd/unit = 200 gpd/unit X 150 units = 30,000 gpd.

This problem illustrates the impact of high rise units versus land space.

2. Find the peak hourly flow in mgd for an 800-acre urban area having the following

features: Domestic flows = 90 gpcd, commercial flows of 15 gpcd, infiltration = 600

gpd/acre, population density of 20 persons per acre and peak hour to average day ratio of

3.0.

Solution:

Calculate the average daily flows for each contributor:

Domestic flows: 90 gpcd X 20 persons per acre X 800-acre = 1.44 mgd

Commercial: 15gpcd X 20 persons per acre X 800-acre = 0.24 mgd

Infiltration: 600 gpd per acre X 800-acre = 0.48 mgd

3. Given a 300 acre housing development with 850 houses (4 persons per house), and

an average annual rainfall of 28 inches, calculate the yearly volume of precipitation over

the area and compare it with the annual sewage flow of 100 gpcd.

Solution:

Yearly volume of 28 inches precipitation, acre-feet per year = 300 acres X 28 inches per

year/12 inches per foot = 700 acre-feet per year

Given Sewage flow = 100 gpcd

Sewage volume, gallons per year would be 850 houses X 4 persons per house X 100 gpcd X

365 days per year = 124,100,000 gallons per year

124,100,000gallons per year / 7.48 gallons per cu.ft. = 16,590,909.09 cu. ft. = 16,590,909.09 cu.

ft. / 43560 sq.ft per acre = 380 acre-feet.

The peak hourly flow = 3.0 peak hour to average day ratio X (1.44+0.24+.48) = 6.48 mgd

4. A mechanically cleaned bar screen has bars 3/8” (5 mm) thick and 1 1/4” clear spaces

between the bars. If the velocity through the bars is 3 feet/sec., determine the approach

velocity, in ft/sec., and the head loss through the screen, in feet.

Solution:

Assume depth = D

Va Aa = VbAb

Va = (Ab/Aa) Vb = (1.25”)(D) / (1.625”) (D) x 3.0 fps = 2.31 fps

HL = [(Vb2 – Va

2) / 2g] (1/0.7) = [(3.0)

2 – (2.31)

2/ (2)(32.17) x (1/0.7) = 0.08 feet.

5. A municipal wastewater plant treats 12.3 MGD. The screenings amount to 1.5 cubic feet

per million gallons treated, and the grit amounts to 4.0 cubic feet per million gallons

treated.

a. Determine the volume in cubic feet of screenings removed per day.

b. Determine the volume in cubic feet of grit removed per day.

Page 18: Pe Civil Eng Problems

18

c. Determine the time in days required to fill a 4’ X 6’ X 6’ solid waste container.

Solution:

a. Volume of screenings = (12.3 MG/Day)(1.5 cu.ft/MG) = 18.5 cu.ft.per day

b. Volume of grit = (12.3MG/day)(4.0cu.ft./MG) = 49.2 cu.ft. per day.

c. Total screenings and grit = 18.5 + 49.2 = 67.7 cu.ft./day

Time = (4’ X 6’ X 6’)(day/67.7 cu.ft.) = 2.13 days

Be able to draw and label the carbon cycle.

Be able to draw and label the nitrogen cycle.

Page 19: Pe Civil Eng Problems

19

Be able to draw and label the sulfur cycle.

Name 6 factors that affect the growth of microorganisms.

1. temperature

2. pH

3. nutrients

4. oxygen supply

5. toxins

6. substrate types

6. A water at elevation 1500 feet(457 mm Hq) is at 20 degrees C. The atmosphere pressure

at Elevation of 1000 feet is 733 mm Hq and at elevation 2200 feet is 706 mm Hq. Determine

the saturation of dissolved oxygen concentration in mg/L?

Solution: Cs = 43.8 mg/L. atmospheres

Elevation 1000 – 733 mm Hq

Elevation 1500 - X

Elevation 2000 – 706 mm Hq.

X = (733 + 706) (0.5) = 719.5 mm

Atmosphere = 719.5 / 760 = 0.9467

Cs = (43.8 mg/L . atm) (0.9467 atm)(0.209) = 8.67 mg/L

7. The high purity oxygen process used in some activated sludge wastewater plants uses

almost pure oxygen gas as an oxygen source instead of air. If the gas is 80% oxygen and

20% nitrogen and is at 1 atm pressure, what is the equilibrium oxygen concentration at 20

degrees C.?

Cs = 43.8 mg/L –atm X Pg

Pg = 0.80 atmospheres

Cs = (43.8 mg/L – atm) (0.80 atm) = 35.0 mg/L

7. An activated sludge plant has anaerobic digesters with sludge gas utilization. The

digesters produce 110,000 cubic feet per day of sludge gas at 10 inches of water column

pressure at 95 degrees F. The gas is to be compressed and stored in a gas dome and used to

fuel internal combustion engines that drive generators to produce electricity. If the

compressed gas is at 30 psig pressure and 95 degrees F, how many cubic feet per day will it

occupy?

Page 20: Pe Civil Eng Problems

20

P1V1 = P2V2

T1 T2

T1 = T2 , so

P1V1 = P2V2

For USCS units:

V1= 110,000 cubic feet

P1 = 14.7 psi + [(10”/12” ft) / 2.31 ft/psi] = 15.06 psi

P2 = 14.7 + 30 psi = 44.7 psi

V2 = P1 = [15.06] (110,000) = 37,100 cubic feet

P2 [ 44,7 ]

Identify the 3 major bacteria types and their temperature ranges used in anaerobic

digestion.

1. Thermophilic bacteria – 50 – 55 degrees C.

2. Mesophilic bacteria – 20 – 40 degrees C

3. Psychrophilic bacteria – 4 – 10 degrees C

Identify the growth phases of a pure culture of bacteria

What is the ideal growth phase for good settling in an activated sludge plant?

The ideal growth phase for good settling in an activated sludge plant is during the endogenous

phase.

Page 21: Pe Civil Eng Problems

21

Label the schematic of a continuous-flow activated-sludge process.

Identify the range of operation for most activated sludge treatment systems in the figure

below.

8. An aerated laboratory vessel is filled with a mixture of substrate and microbial cells. At

time zero, the substrate concentration (S) is 150 mg/L and the cell concentration (X) is 1500

mg/L. The biochemical reaction is pseudo-first order, and the rate constant K is 0.40 L/

(gm cells)(hr). After 6 hours of aeration, the substrate concentration (S) is 4 mg/L, and the

Page 22: Pe Civil Eng Problems

22

cell concentration (X) is 1590 mg/L. Using the specific rate of substrate utilization

equation, (1/X)(dS/dt) = KS, determine: a. The rate of substrate utilization, dS/dt, at time

zero in mg/(L)(hr). b. The rate of substrate utilization, dS/dt, at time equals 6 hours in

mg?(L)(hr).

1 dS = KS where S is substrate concentration, t is time, X is cell concentration

X dt

Or

dS = XKS

dt

a. dS = (1500 mg ) ( grams ) (0.40L) (150 mg) = 90 mg/L- hr

dt L 1000 mg) (gm-hr) ( L )

b. ds = [1591 mg] (grams ) (0.40 L) (4 mg) = 2.5 mg/L – hr.

dt L 1000 mg) (gm-hr) ( L)

9. A municipal effluent has a BOD5 of 10 mg/L and K1(base e) is 0.23 day-1

. Determine:

a). The ultimate first stage BOD in mg/L, Lo.

b). The ratio, y/Lo as a fraction and is a percent.

Solution:

a). y = Lo (1 – e-kt

)

10 = Lo (1- e-0.23x5

)

Lo = 14.6 mg/L

b). y/Lo = 10/14.6 = 68.5%

Label the diagr6am of population dynamics in anaerobic digestion.

What causes the pH drop in anaerobic digestion.

An excess of organic matter is fed to a digester will cause acid formers to rapidly process the

food resulting in excessive organic acids while the methane formers are unable to metabolize the

organic acids fast enough resulting in a pH drop.

Explain symbiosis as it applies to bacteria and algae in oxidation ponds.

Symbiosis is the relationship of two or more species that live together for mutual benefit such as bacteria that metabolize organic matter releasing nitrogen and phosphorous and carbon

Page 23: Pe Civil Eng Problems

23

dioxide for the benefit of algae that use these compounds with energy from sunlight for

synthesis, releasing oxygen for the benefit of the bacteria.

Define Facultative Stabilization Ponds. Facultative stabilization ponds are anaerobic at the bottom of the ponds while the surface is

aerobic.

10. Calculate the pounds of BOD contributed to a sewage plant each day from a

population of 100000 with an average of 0.20 pounds of BOD per person per day.

Solution: 100000 X 0.20 = 20,000 pounds of BOD per day.

11. Calculate the pounds of suspended solids contributed to a sewage plant each day from

a population of 100000 with an average of 0.24 pounds of S.S. per person per day.

Solution: 100000 X 0.24 = 24,000 pounds of Suspended solids

12. Given the flow of 10 mgd with a raw BOD or suspended solids level of 250 mg/L,

determine the pounds of BOD or suspended solids level contributed to the sewage plant.

Solution: 10 mgd X 8.34#/MG X 250 ppm = 20850 pounds

13. If domestic wastewater contains 0.24 pounds of suspended solids and 0.20 pounds of

BOD per 120 gallons per day per capita, what is the BOD equivalent population for an

industry that discharges 0.10 mgd of wastewater with an average BOD of 450 mg/L? What

is the hydraulic equivalent population of this wastewater?

Solution:

BOD equivalent population = 0.10 MG X 450 mg/L X 8.34 lb/MG / 0.20 lb/person = 1900

Hydraulic equivalent population = 100000 gal/day/120 gal/person = 830

Define the following terms:

Carbohydrates – are hydrates of carbon with the empirical formula CnH2nOn.

Monosaccharide – simplest carbohydrate unit such as glucose.

Disaccharides – are composed of two monosaccharide units such as sucrose.

Polysaccharides – are long chains of monosaccharides such as cellulose, starch and glycogen.

Cellulose – is the common polysaccharide in wood, cotton, paper, and plant tissues.

Starches – are primary nutrient polysaccharides for plant growth and are abundant in potatoes,

rice, wheat, corn and other plant forms.

Proteins – in simple form are long-chain molecules composed of amino acids connected by

peptide bonds and are important in both the structure ( muscle tissue) and dynamic aspects

(enzymes) of living matter.

Lipids – form the bulk of organic matter of living cells which are soluble in varying degrees in

organic solvents while being only sparingly soluble in water.

Page 24: Pe Civil Eng Problems

24

Be able to label the parts of a trickling filter

Label the parts of the schematic diagram of the trickling filter process.

14. Calculate the total volume of media required for a trickling filter if 35 lbs/1000 cu.

Ft/day of BOD is applied at a flow of 1.2 mgd at 315 mg/L BOD and 35% BOD removal is

expected through the filter.

Solution:

1.2 MGD X 8.34 lbs/gallon X 315 mg/L X .65 BOD remaining / 0.035 lbs/cu ft. = 58,500 cu. Ft.

15. Calculate the % BOD removal at 20 degrees C using the NRC formula for a single-

stage trickling filter given the following information:

NRC formula = E20 = 100 / 1 + 0.0561 (w/VF)0.5

w/V = BOD loading, 23.5 lbs/1000 cu.ft./day

F, recirculation factor = 1.36

E20 = 100/ 1 + 0.0561(23.5/1.36)0.5

= 81.1%

16. Calculate the plant efficiency if the primary sedimentation removes 35% BOD, first

stage trickling filter removes 70% BOD, and the second stage trickling filter removes 60%

BOD.

Solution:

E = 100 – 100[(1-0.35)(1-0.70)(1-0.60)] = 92.2%

Page 25: Pe Civil Eng Problems

25

Label the parts of the continuous-flow activated sludge process.

Name five activation-sludge processes.

1. Step Aeration

2. Conventional

3. Contact Stabilization

4. Extended Aeration

5. High-Purity Oxygen

Name two ways of expressing BOD loadings in an activated-sludge process.

1. BOD/day/1000 cu. Ft

2. BOD/day/lb of MLSS (mixed liquor suspended solids) or MLVSS (mixed liquor volatile

suspended solids)

17. Convert the average BOD concentration of 200 mg/L into units of pounds per 1000 cu.

ft/day.

Solution:

MGD X 8.34#/gal X mg/l = pounds/day

1 MGD X 8.34 X 200 = 1668 pounds/ one million gallons/day or 133689 cu. Ft/day

1668/133689 = .0124767 pounds/cu.ft/day

0.01247 X 1000 cu. Ft. = 12.47 pounds/1000 cubic feet/day

18. If the BOD loading of 12.5 pounds/1000 cu.ft is aerated for a period 6 hours, determine

what the BOD loading becomes in terms of pounds/1000 cubic feet/day?

12.5lb/1000 cu. Feet X 24/6.0 hr = 50 lb/1000 cu.ft./day.

19. Calculate the BOD loading in terms of lb/day/lb MLSS of an activated sludge if the

settled wastewater flow = 3.67 mgd, the influent BOD = 128 mg/L, the MLSS in aeration

tank = 2350 mg/L, the aeration tank volume = 0.898 million gallons.

Solution: BOD load = 3.67 X 128 mg/L X 8.34 = 3920 lb/day

MLSS in aeration tank = 0.898 MG X 2350 mg/L X 8.34 = 17,600 lb.

BOD loading = 3920/17600 = 0.22 lb/day/lb of MLSS

Page 26: Pe Civil Eng Problems

26

20. Determine the sludge age in days when given the volume and suspended solids in

aeration tank, volume and suspended solids in effluent, and volume and suspended solids in

the waste sludge.

Solution:

Sluge age = MLSS X V/SSe X Qe + SSw X Qw

where MLSS = mixed liquor suspended solids, mg/L

V = volume of aeration tank, million gallons

SSe = suspended solids in effluent, mg/L

SSw = suspended solids in waste sludge, mg/l

Qe = quantity of effluent wastewater, mgd

Qw = quantity of waste sludge, mgd

Example:

MLSS = 2350 mg/L

V = 120,000 cu.ft = 0.898 million gallons

SSe = 26 mg/L

Qe = 3.67 mgd

SSw = 11,000 mg/L

Qw = 0.0189 mgd

Sludge age = mean cell residence time = 2350 X 0.898 / 26 X 3.67 + 11,000 X 0.0189 = 7.0

days.

How does step-aeration and tapered aeration differ from the conventional activated sludge

process?

In step aeration the influent load is introduced at several points along the tank length while the

tapered aeration attempts to supply air to match oxygen demand along the length of the tank.

Step loading provides more uniform oxygen demand for an evenly distributed air supply.

Identify the step-aeration activated sludge process diagram.

Page 27: Pe Civil Eng Problems

27

Identify the Contact Stabilization Process without primary sedimentation.

Identify the Extended Aeration Process without primary sedimentation.

Identify 6 biological factors that can adversely affect settleability of activated sludge.

1. Species of dominant microorganisms (filamentous)

2. Ineffective biological flocculation

3. Denitrification in final clarifier (floating solids)

4. Excessive volumetric and food/microorganisms loadings

5. Mixed-liquor suspended-solids concentration

6. Unsteady-state conditions (nonuniform feed rate and discontinuous wasting of excess

activated sludge).

Identify 5 chemical factors that can adversely affect settleability of activated sludge.

1. Lack of nutrients

2. Presence of toxins

3. Kinds of organic matter

4. Insufficient aeration

5. Low temperature

Identify 2 physical factors that can adversely affect settleability of activated sludge.

1. Excessive agitation during aeration resulting in shearing of floc.

2. Ineffective final clarification: inadequate rate of return sludge, excessive overflow rate or

solids loading, or hydraulic turbulence.

Page 28: Pe Civil Eng Problems

28

21. Calculate the specific gravity of solid matter in a sludge based on equation 13.1 as

follows:

Eq. 13.1: Ws/Ssy = Wf/Sfy + Wv/Svy, where Ws = weight of dry solids, lb., Ss = specific

gravity of solids, y = unit weight of water, lb/cu.ft, Wf = weight of fixed solids (nonvolatile), lb.,

Sf = specific gravity of fixed solids, Wv = weight of volatile solids, lb., Sv=specific gravity of

volatile solids.

Example: Consider a waste biological sludge of 10% solids with a volatile fraction of 70%.

Their specific gravity can be estimated using the above equation by assuming values of 2.5 for

the fixed matter and 1.0 for the volatile residue.

Solution: 1.00/Ss = 0.30/2.5 + 0.70/1.0 = 0.82

Ss = 1/0.82 = 1.22

22. Calculate the specific gravity of the wet sludge by Eq. 13.2 as follows:

S = Ww + Ws/(Ww/1.00) + (Ws/Ss), where S = specific gravity of wet sludge; Ww = weight of

water, lb; Ws = weight of dry solids, lb.; Ss = specific gravity of dry solids

Example: Determine the specific gravity of a wet sludge of 10% solids.

S = ___90 + 10________ = 1.02

(90/1.00) + (10/1.22)

23. Estimate the quantity of sludge produced by a trickling filter plant treating 1.0 mgd of

domestic wastewater. Assume a suspended solids concentration of 220 mg/L in the raw

wastewater, a solids content in the sludge equivalent to 90% removal and a sludge of 5.0%

concentration withdrawn from the settling tanks.

Solution: Solids in the sludge = 1.0 X 8.34 X 220 X 0.90 = 1650 lbs/day

Volume of Sludge (using Eq. 13.3) = 1650/0.05 X 62.4 = 530 cu.ft/day

24. Calculate the total dry solids in lbs/day in primary settling with 50% settling of the

solids for a wastewater plant treating 1 mgd of 200 mg/L suspended solids in unsettled

wastewater.

Solution: Wsp = f X SS X Q X 8.34, where Wsp = primary solids, lb of dry weight/day; f =

fraction of suspended solids removed in primary settling; SS = suspended solids in unsettled

wastewater, mg/L; Q = daily wastewater, mgd; 8.23 = conversion factor, lb/mg per mg/L

Ws = 0.50 X 200 X 1.0 X 8.34 = 834 lbs/day

25. Calculate the total dry solids from an activated sludge process without primary

sedimentation, in lbs/day of dry weight from a wastewater plant treating 1 mgd with a FM

ratio of 0.05 and an influent BOD of 250 mg/L.

Page 29: Pe Civil Eng Problems

29

Was, total dry solids from activated-sludge processes without primary sedimentation, lb/dayof

dry weight = 2.0K X BOD X Q X 8.34, where K is the factor determined from figure 13.1

multiplied by 2.0.

Was = 2.0(0.3) X 250 mg/L X 1.0 MGD X 8.34#/gallons = 1251 lbs dry weight solids/day

26. A municipal wastewater plant effluent sample was filtered through a crucible and

filter mat. The crucible plus the mat had a dry (tare) weight of 17.8216 grams. After

filtration, the crucible mat and residue weighed 17.8374 gm. After weighing the crucible,

mat, and residue were ignited at 600 degrees C. The crucible, mat, and ash weighed

17.8258 grams. If the sample was 50 ml. determine the suspended solids, the volatile solids,

the fixed solids, and the percent volatile suspended solids.

Solution:

17.8374

-17.8216

0.0158 gm = 15.8 mg

TSS = (15.8 mg/50 ml)(1000 ml/L) = 316 mg/L

17,8258

-17.8216

0.0042 gm = 4.2 mg

FSS = (4.2/50 ml) (1000 mg/L) = 84 mg/L

VSS = TSS – FSS = 316 – 84 = 232 mg/L

Percent VSS (232/316) 100% = 73.4%

Label the parts of a gravity sludge thickener

27. Determine the diameter of a gravity thickener based on a solids loading of 10

lb/sq.ft/day and 1130 pounds of solids per day.

Solution: Tank area required = 1130 lbs. per day / 10 lbs. per sq ft per day = 113 sq.ft.

Use equation: П D2/4 = area = 113 sq. ft : Solve diameter of tank.

П D2 = 452

D2 = 143.87

D = 12.0 ft.

28. Determine the gallons per day of 1130 pounds per day of applied sludge at a

concentration of 4.5%.

Solution: Volume of applied sludge = 1130 lbs. per day / (0.045 sludge conc.X 62.4 lbs. per cu.

ft of water) = 402 cu.ft of wet sludge.

402 cu. ft of wet sludge X 7.48 gallons per cu. ft. = 3007 gallons per day of sludge.

Page 30: Pe Civil Eng Problems

30

29. Determine the gpd of dilution water required to obtain an overflow rate of 400

gpd/sq.ft if 3007 gpd of applied sludge is sent to the gravity thickener and the area of the

thickener is 113 square feet.

Solution: Overflow rate of applied sludge = 3007 gpd/113 sq.ft. = 27 gpd/sq.ft.

Dilution flow required to attain 400 gpd/sq.ft = (400-27) 113 = 42,149 gpd ~ 42,000 gpd.

30. Determine the solids retention time in a gravity thickener with a surface area of 113 sq.

feet from a trickling filter plant that contains 1130 pounds per day of solids assuming an

underflow of 8.0% solids concentration and 95% solids capture at a sludge depth of 3.0

feet.

Solution: Volume of thickened sludge = 1130 lbs per day X 0.95 / (8.0/100) 62.4 lbs. per

cu. ft of water = 215 cu. Ft/ day

Solids retention time = 3 feet depth X 113 sq. ft. area X 24 hours per day / 215 cu.

ft per day = 38 hours.

31. A single-stage anaerobic digester has a capacity of 13,800 cu.ft., of which 10,600 cu.ft is

below the landing brackets. The average raw-waste sludge solids fed to the digesters are

580 pounds of solids/day. Calculate the digester loading in pounds of volatile solids fed per

cubic feet capacity below the landing brackets/day. Assume that 70% of the solids are

volatile.

Solution:

Volatile solids loading = 0.70 volatile solids X 580 lbs. of volatile solids per day / 10,600 cu.ft

digester volume = 0.038 lb of volatile solids fed / cu.ft./day

32. Calculate digester capacity using the following data based on the following equation

and data:

Vtotal = [(V1 + V2)/2 X T1] + [V2 X T2] where V = total digester capacity, cu.ft; V1 =

volume of average daily raw-sludge feed, cuft/day; V2 = volume of daily digested sludge

accumulation in tank, cu.ft/day; T1 = period required for digestion, days (approx. 25 days

at temperature of 85 – 95 degrees F; T2 = period of digested sludge storage, days (normally

30-120 days).

Problem Data:

Average daily raw-sludge solids = 580 pounds

Raw-sludge moisture content = 96%

Digestion period = 30 days

Solids reduction during digestion = 45%

Digested-sludge moisture content = 94%

Digested-sludge storage required = 90 days

Solution:

V1, volume of avg daily feed solids = 580avg daily raw solids/0.04 dry solids X62.4 lbs. per cu.

ft = 232 cu.ft.

V2, volume of daily digested sludge, cu. ft per day = 0.55 X 580/ 0.06 X 62.4 = 85 cu.ft per day

Solution:

Vtotal = [(232 cu ft + 85 cu ft)/ 2 (30 days digestion)] + [(85 cu. ft X 90 days digestion)] =

12,400 cu.ft. of digester volume

Name three important parameters in nitrification kinetics.

Temperature, pH, and dissolved-oxygen concentration

Page 31: Pe Civil Eng Problems

31

What is the minimum dissolved oxygen level recommended in practice for nitrification to

prevent reduced nitrification through the aeration tank?

2.0 mg/L is recommended in practice.

Why is a long sludge age required in nitrification to prevent excessive loss of viable

bacteria in continuous flow aeration systems?

A long sludge age is required because the growth rate of viable bacteria must be rapid enough to

replace microbes lost through sludge wasting and washout in the plant effluent.

33. Calculate the Mean Cell Residence Time in activated-sludge operation using equation:

Equation: MLVSS X V/SSe X Qe + SSw X Qw

Given: MLVSS = 1400

V = Volume of aeration tank = 1.5 million gallons

SSe = Effluent suspended solids = 20 mg/L

Qe = design flow = 5.0 mgd

SSw = suspended solids in waste sludge, mg/l =8200 mg/L

Qw =quantity of waste sludge, mgd= .026 mgd

Solution:

(1400 mlvss X 1.5 MG) / [(20 mg/L X 5.0 MGD) + 8200 mg/L X 0.026 MGD)] = 6.7 days

34. Calculate the Mean Cell Risidence Time for Nitrification given equation at 17 degrees

C.

Equation: µN = 0.47e0.098(T-15)

where µN = maximum specific growth rate of Nitrosomonas, d-1

e = base of Napierian logarithms, 2.718

T = temperature, centigrade

Solution:

µN = 0.47(e)0.098(17-15)

= 0.572 days

35. A rapid sand filter has a sand bed 30 inches in depth. Pertinent data are: specific

gravity of the sand = 2.65, shape factor (Ø) = 0.75, porosity (ε) = 0.41, filtration rate = 2.25

gpm/sq.ft, and operating temperature = 50 degrees F. The sieve analysis of the sand is

given in attached table. Determine the head loss for the clean filter bed in a stratified

condition using the Rose equation:

HL= (1.067/Ø) (D/g) (Va2/ε

4) (Σ CDX/d). D=feet, Va = ft/sec, g = gravity or 32.17 ft/sec

2

Page 32: Pe Civil Eng Problems

32

HL = (1.067/0.75)(2.5ft/32.17 ft)(0.0050133 ft/sec)2 X (1/0.41)

4 (35,700.0 ft

-1) = 3.51 feet

36. Filter Loading Rate = gpm/sq ft filter area Example 1: A rapid sand filter is 10 feet wide and 15 feet long. If the flow through the filter is

450,000 gpd, what is the filter loading rate in gallons per minute per square foot?

Filter loading rate = gpm/sq.ft filter area

Gpm/sq.ft = 450000gallons/1440 minutes/ (10 ft X 15 ft) = 2.08 gpm/sq.ft.

37. Filter loading rate = Inches of water fall / minute

Example 2: How many inches drop per minute corresponds to a filter loading rate of 2.6 gpm/sq.

ft?

Inches drop/minute = 2.6 gpm/7.48 gals/ cu.ft. (12 inches/foot) = 4.17 inches/minute

38. What is the filter loading rate in gpm/sq. ft. of a sand filter in which the water level

dropped 15 inches in 3minutes after the influent valve was closed?

15 inches/3 minutes = 5 inches per minute

5 in/min / 12 (7.48 gals/cu.ft) = 3.12 gpm/sq.ft.

39. A rapid sand filter is 15 feet wide and 40 feet long. If the flow through the filter is 1.6

mgd, what is the filter loading rate in gpm/sq. ft?

1,600,000/1440 minutes/day = 1111 gpm

Filter loading rate = gpm/sq ft area = 1111 gpm / 15 X 40 = 1.85 gpm/sq.ft.

40. Filter backwash rate = gpm/sq.ft.area

A rapid sand filter is 10 feet wide and 16 feet long. If backwash water is flowing upward at

a rate of 4,950,000 gpd, what is the backwash rate in gallons per minute per square foot?

Filter backwash rate = gpm/sq.ft area = 4,950,000 / 1440 minutes/day / 10 X 16 = 21.48

gpm/sq.ft.

41. A mixed media filter is 25 feet wide and 32 feet long. If the filter receives a backwash

flow of 17,300,000 gpd, what is the filter backwash flow rate in gpm/sq.ft.

Filter backwash rate = gpm/sq.ft area = 17,300,000 / 1440 min/day / 25 X 32 = 15.02 gpm/sq.ft.

42. How many inches rise per minute corresponds to a filter backwash rate of 18

gpm/sq.ft.?

18 gpm/sq.ft /7.48gal/cu.ft (12 inches/foot) = 28.9 inches per minute.

Page 33: Pe Civil Eng Problems

33

43. A rapid sand filter is 15 ft wide and 30 feet long. If the backwash water flow rate is 12

mgd, determine the filter backwash rate in gpm/sq.ft.

Filter backwash rate, gpm/sq.ft = 12,000,000/1440 minutes/day / 15 ft X 30 ft = 18.52 gpm/sq.ft.

44. During backwashing of a rapid sand filter, the operator measured the backwash flow

rate to be 27 inches/minute. Express this filter backwash rate in gallons per minute per

square foot.

Gpm/sq. ft. = 27 inches/min / 12 inches/foot X 7.48 gal/cu.ft = 16.88 gpm/sq.ft.

45. Determine the design bed volume , cu.ft., for a flow rate of 40,000 gpd with an

experimental pilot column result of 1.67 BV/hour.

BV, cu. ft = Q/QB , where design flow rate = Q; QB = Breakthrough volume from an

experimental pilot column

Design BV, cu. ft. = 40000 X hour X _cu.ft.___ = 133 cu. ft.

24 hrs 1.67 7.48gal

46. If the mass of carbon = 25 lb/cu.ft, how much is required for the design bed volume of

133 cu. ft?

M, lb = (BV)(ρs) where BV, cu. ft and ρs = mass of carbon in lb/cu. ft.

M, lb = (133)(25) = 3330 lb.

47. If the volume treated at the allowable breakthrough was 549.5 gallons, what was the

number of gallons per pound treated if the pilot column’s mass of carbon was 6.564

pounds?

Vb = Vb/M = 549.5/ 6.564 = 83.71 gal/lb.

48. How many pounds of carbon per hour were exhausted in treating 40,000 gallons per

day if each pound of carbon required 83.71 gallons?

Mt = Q/Vb, Mt = pounds exhausted per hour; Q = flow rate; Vb = gallons per pound treated

Mt = 40000/24 hr (lb/83.71) = 19.91 lb/hour

49. How many days would it take for the breakthrough time if the bed contained 3330

pounds of carbon and the bed was exhausted at the rate of 19.91 pounds per hour?

T = M/Mt = (3330) (hr/19.91 lb) = 167 hours or 6.96 days.

50. A pilot column breakthrough test has been performed. Pertinent design data are

inside diameter = 3.75 inches, length = 41 inches, mass of carbon = 2980 grams or 6.564

lbs., liquid flow rate = 17.42 liters per hour or 1.00 gpm/sq.ft., and packed carbon density =

25 lbs/cu.ft (401 gm/liter. The breakthrough of the carbon occurred after 295 gallons had

passed through the column. Determine: 1). the liquid flowrate in bed volumes per hour

and 2. the volume of liquid treated per unit mass of carbon ----in other words, the gal/lb at

an allowable breakthrough amount.

1. BV, = area X length X (cu ft/ 1728 cu.in.)(7.48 gals/cu.ft)(3.785liters/gal) = liters

BV = (π(3.75 in)2/4 X 41)(cu.ft/1728 cu.in)(7.48/cu.ft)(3.785 liters/gal) = 7.4193 liters

BV = Q/Qb or Qb = Q/BV = 17.42liter/hr/7.4193 liters = 2.35 BV/hour

2. Vb = 295 gal/6.564 lb = 44.9 gal/lb.

Page 34: Pe Civil Eng Problems

34

51. A water is to be softened for an industrial water supply. The calcium content is 107

mg/L and the magnesium content is 18 mg/L. The allowable breakthrough, Ca, is 5%,

where Co is the hardness of the untreated water. A pilot column 4 inches in diameter and

containing 3 feet of resin has been operated at a flowrate of 0.59 gal/hour to obtain

breakthrough data. The resin has a specific weight of 44lb/cu.ft., and the design flowrate is

150,000 gal/day. Using the scale-up design approach, determine:

1. The pilot column volume(cu.ft)

2. The pilot column mass(lb)

3. The pilot column BV/hour

4. The full scale up volume(cu.ft)

5. The full scale up mass of resin(lb)

6. The amount of millequivalents of total hardness

7. If 740 gallons is obtained from the breakthrough curve, how many gallons per pound

are obtained for the pilot column?

8. How many pounds per hour would the full scale column unit operate based on the pilot

scale unit’s gallons per pound rate.

9. What is the design life of the column?

1. V = (π(4 in/12 in/ft)]2/4) (3 feet) = 0.2618 cu. ft.

2. M = (0.2618 cu.ft)[44 lb/cu.ft.] = 11.519 lb.

3. Qb = (0.59 gal/hr)(BV/0.2618 cu.ft.)(cu.ft/7.48 gal) = 0.301 BV/hr.

4. BV = Q/Qb = (150000 gal/day)(1 day/24 hrs)(1 hr/0.301 BV)(cu.ft/7.48 gal) = 2780 cu. ft.

5. lb resin = (2780 cu. ft.)(44 lb/cu.ft) = 122000 lb.

6. Total hardness = 107/20 + 18/12 = 6.85 meq/l

7. Vb = 740 gal/11.519 lb = 64.24 gal/lb.

8. Mt = (150000 gal/day) (1 day/24 hours)/ Vb = 97.3 lb/hour

9. t = 122000 lb/ 97.3 lb/hour = 1250 hours, or 52.1 days

52. A soluble organic industrial wastewater with a COD of 2590 mg/L is to be treated by

completely mixed activated sludge plant. A treatability study has been made, and the

following data obtained: Design influent flow = 4.2 MGD, MLSS = 3000 mg/L, MLVSS =

86.7% of the MLSS, K = 0.236 liter/ (gm MLVss-hr), reaction is pseudo-first order, non-

degradable COD = 16 mg/L, and effluent COD = 150 mg/L, Determine:

a. Required reaction time, Ө, hours

b. Required Reactor basin volume, cubic feet.

Solution:

a. Si = 2590 – 16 = 2574 mg/L

St = 150 – 16 = 134 mg/l

MLVSS = (0.867)(3000) = 2600 mg/L: = 2.60 gm/L

Ө = Si – St = 2574 - 134 = 28.7 hours (Required Reaction Time)

K(MLVSS)(St) (0.236)(2.60)(134)

b. V = QӨ = (4,200,000 gals)(29.7 hrs.)(ft3/7.48gal) = 695,000 ft

3

Page 35: Pe Civil Eng Problems

35

Page 36: Pe Civil Eng Problems
Page 37: Pe Civil Eng Problems
Page 38: Pe Civil Eng Problems
Page 39: Pe Civil Eng Problems
Page 40: Pe Civil Eng Problems
Page 41: Pe Civil Eng Problems
Page 42: Pe Civil Eng Problems
Page 43: Pe Civil Eng Problems
Page 44: Pe Civil Eng Problems